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Unformatted text preview: 0 P1 1 This is a quadratic equation in the variable R1. The solution can be found by using the
quadratic formula (see Appendix C.4): R1 = = V2 − 2 R2 −
± P
1 2 V2 2 2 R2 − − 4 R2 P
1 2 (120.0 V )2 − 2 (144 Ω ) −
±
23.4 W 2 (120.0 V )2 2
2 (144 Ω ) − − 4 (144 Ω )
23.4 W 2 = 85.9 Ω and 242 Ω
______________________________________________________________________________
47. SSM REASONING The resistance of one of the wires in the extension cord is given by . Equation 20.3: R = ρ L / A , where the resistivity of copper is ρ = 1.72 × 10−8 Ω m,
according to Table 20.1. Since the two wires in the cord are in series with each other, their
total resistance is Rcord = Rwire 1 + Rwire 2 = 2 ρ L / A . Once we find the equivalent resistance
of the entire circuit (extension cord + trimmer), Ohm's law can be used to find the voltage
applied to the trimmer. 1082 ELECTRIC CIRCUITS SOLUTION
a. The resistance of the extension cord is Rcord = 2 ρ L 2(1.72 ×10 –8 Ω ⋅ m)(46 m)
=
= 1.2 Ω
A
1.3 ×10 –6 m2 b. The total resistance of the circuit (cord + trimmer) is, since the two are in series, Rs = 1.2 Ω + 15.0 Ω = 16.2 Ω
Therefore from Ohm's law (Equation 20.2: V = IR ), the current in the circuit is I= V 120 V
=
= 7.4 A
Rs 16.2 Ω Finally, the voltage applied to the trimmer alone is (again using Ohm's law),
Vtrimmer = (7.4 A)(15.0 Ω) = 110 V ______________________________________________________________________________
48. REASONING The answer is not 340 W + 240 W = 580 W. The reason is that each heater
contributes resistance to the circuit when they are connected in series across the battery.
For a series connection, the resistances add together to give the equivalent total resistance,
according to Equation 20.16. Thus, the total resistance is greater than the resistance of
either heater. The greater resistance means that the current from the battery is less than
2
when either heater is present by itself. Since the power for each heater is P = I R,
according to Equation 20.6b, the smaller current means that the power delivered to an
individual heater is less when both are connected than when that heater is connected alone.
We approach this problem by remembering that the total power delivered to the series
combination of the heaters is the power delivered to the equivalent series resistance.
SOLUTION Let the resistances of the two heaters be R1 and R2. Correspondingly, the
powers delivered to the heaters when each is connected alone to the battery are P1 and P2.
For the series connection, the equivalent total resistance is R1 + R2, according to
Equation 20.16. Using Equation 20.6c, we can write the total power delivered to this
equivalent resistance as
V2
P=
(1)
R1 + R 2 But according to Equation 20.6c, as applied to the situations when each heater is connected
by itself to the battery, we have Chapter 20 Problems 1083 P1 = V2
R1 or R1 = V2
P1 (2) P2 = V2
R2 or R2 = V2
P2 (3) Substituting Equations (2) and (3) into Equation (1) gives
P= V2 = 1 = P1 P2 = (340 W )(240 W ) = 140 W
1
1
P1 + P2
340 W + 240 W
V2 V2
+
+
P1 P2
P1
P2
______________________________________________________________________________ 49. REASONING Ohm’s law provides the basis for our solution. We will use it to express the
current from the battery when both resistors are connected and when only one resistor at a
time is connected. When both resistors are connected, we will use Ohm’s law with the series
equivalent resistance, which is R1 + R2, according to Equation 20.16. The problem statement
gives values for amounts by which the current increases when one or the other resistor is
removed. Thus, we will focus attention on the difference between the currents given by
Ohm’s law.
SOLUTION When R2 is removed, leaving only R1 connected, the current increases by
0.20 A. In this case, using Ohm’s law to express the currents, we have VR2
V
V
–
=
= 0.20 A
R1 + R2
R1
R1 (R1 + R 2 )
14 244
4
3 1442443
Current given by
Ohm' s law when
only R 1 is present Current given by
Ohm' s law when
R1 and R 2 are present (1) When R1 is removed, leaving only R2 connected, the current increases by 0.10 A. In this
case, using Ohm’s law to express the currents, we have
VR1
V
V
–
=
= 0.10 A
R2
R1 + R2
R 2 (R1 + R2 )
14 244 1442443
4
3
Current given by
Ohm' s law when
only R 2 is present Current given by
Ohm' s law when
R1 and R 2 are present Multiplying Equation (1) and Equation (2), we obtain V R2 V R1 = ( 0.20 A )( 0.10 A ) R1 ( R1 + R2 ) R2 ( R1 + R2 ) (2) 1084 ELECTRIC CIRCUITS Simplifying this result algebraically shows that
V 2 ( R1 + R2 ) 2 = ( 0.20 A )( 0.10 A ) or V
=
R1 + R2 ( 0.20 A )( 0.10 A ) = 0.14 A (3) a. Using the result for V/(R1 + R2) from Equation (3) to substitute into Equation (1) gives V
– 0.14 A=0.20 A
R1 or R1 = V
12 V
=
= 35 Ω
0.20 A+0.14 A 0.20 A+0.14 A b. Using the result for V/(R1 + R2) from Equation (3) to substitute into Equation (2) gives V
V
12 V
1
=
= 5.0 × 10 Ω
– 0.14 A=0.10 A
or
R2 =
R2
0.10 A+0.14 A 0.10 A+0.14 A
__________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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